integral of (x^2)/(x^2+x+12)

Lösung

\int \frac{x ^{2}}{x ^{2} + x + 12} d x

- \frac{1}{2} ln 4 x^{2} + 4 x + 48 - \frac{23}{47} arctan (\frac{1}{47} (2 x + 1)) + x + C

Schritte zur Lösung

\int \frac{x ^{2}}{x ^{2} + x + 12} d x

\frac{x ^{2}}{x ^{2} + x + 12} = \frac{- x - 12}{x ^{2} + x + 12} + 1

= \int \frac{- x - 12}{x ^{2} + x + 12} + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{- x - 12}{x ^{2} + x + 12} d x + \int 1 d x

\int \frac{- x - 12}{x ^{2} + x + 12} d x = - \frac{1}{2} ln 4 x^{2} + 4 x + 48 - \frac{23}{47} arctan (\frac{1}{47} (2 x + 1))

\int 1 d x = x

= - \frac{1}{2} ln 4 x^{2} + 4 x + 48 - \frac{23}{47} arctan (\frac{1}{47} (2 x + 1)) + x

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln 4 x^{2} + 4 x + 48 - \frac{23}{47} arctan (\frac{1}{47} (2 x + 1)) + x + C

x y \frac{d y}{d x} = 7

\int_{0}^{1} sin (π t + \frac{π}{4}) d t

y^{^{'''}} + 16 y^{^{'}} = 0

y^{^{'}} = \frac{( x - 1 )}{y}

\frac{d}{d x} (\frac{1}{3 ^{(- x^{2} - y^{2})} + 1})